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Let $\Gamma$ be a countable group and let $\Lambda_1,\Lambda_2<\Gamma$ be subgroups. We say that $\Lambda_1$ is amenable relative to $\Lambda_2$ if the action of $\Lambda_1$ on $\Gamma/\Lambda_2$ by left translation admits an invariant mean.

Let now $\Sigma<\Lambda<\Gamma$ be subgroups and let $G<N_\Gamma(\Sigma)$ be a subgroup of the normalizer of $\Sigma$ in $\Gamma$. Assume that $G$ is amenable relative to $\Lambda$.

Question: Is $\Sigma G$ still amenable relative to $\Lambda$ ?

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  • $\begingroup$ I guess $\Sigma G$ is not (generally) amenable relative to $\Lambda$. Take for example $G=\{e\}$, the trivial group, then $\Sigma$ has to be amenable relative to $\Lambda$. I notice that my counterexample is not a complete proof, but I hope it sheds some light to the problem. $\endgroup$
    – user23860
    Commented Sep 26, 2013 at 14:19
  • $\begingroup$ @Vahid: Although it is not explicitly stated, I'm interested in the case when $G$ is nontrivial. $\endgroup$
    – bemihai
    Commented Sep 26, 2013 at 14:25
  • $\begingroup$ Actually it doesn't matter, because if $\Sigma G$ is amenable relative to $\Lambda$, then $\Sigma$ (as a subgroup of $\Sigma G$) has to be amenable relative to $\Lambda$. $\endgroup$
    – user23860
    Commented Sep 26, 2013 at 14:34

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Here is a counterexample. Take $\Delta$ a non-amenable group, and set $\Gamma = (\mathbb Z/2 \mathbb Z) \ltimes (\Delta \times \Delta)$, where the action is via the flip automorphism on $\Delta \times \Delta$. Let $\Sigma = \Lambda$ be the first copy of $\Delta$, and let $G$ be the second copy of $\Delta$.

Then $G$ is conjugate to $\Lambda$ and hence $\Gamma / \Lambda$ has a $G$-fixed point, and in particular, has a $G$-invariant mean. On the other hand, $\Gamma / \Lambda$ has two $\Sigma G$-orbits and neither can have an invariant mean since both witness the non-amenability of $\Delta$.

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